I have read that sine and cosine can be represented as power series. Power series, as I understand them, are infinite series that can be represented as:
$\sum_{j=0}^{\infty} a_j (x-x_0)^j$
where $a_j$ is not dependent on $x$. However, I'm not sure how the representations of sine and cosine satisfy this requirement. For example, the formula for $sin(x)$ is given as:
$\sum_{j=0}^{\infty} \frac{(-1)^j}{(2j+1)!} x^{2j+1}$
How can I rewrite this in the form, $\sum_{j=0}^{\infty} a_j (x-x_0)^j$?
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$\begingroup$@Git Gud wrote: $$\sum_{j=0}^{\infty} \frac{(-1)^j}{(2j+1)!} x^{2j+1}=\sum_{j=0}^{\infty} \underbrace{\frac{(-1)^j}{(2j+1)!}}_{\displaystyle a_{2j+1}} (x-\underbrace{0}_{\displaystyle x_0})^{2j+1}$$
Going a bit further, set $a_j=\begin{cases} \frac{(-1)^{(j-1)/2}}{j!} & j\textrm{ odd}\\ 0 & j\textrm{even}\end{cases}$. Now $\sum_{j=0}^{\infty} \frac{(-1)^j}{(2j+1)!} x^{2j+1}=\sum_{j=0}^{\infty} a_j (x-0)^j$.
$\endgroup$ 0 $\begingroup$The first series you gave converges for all $x$, and is of the right shape, so there is no need to rewrite. (The fact the even powers of $x$ were not mentioned doesn't matter: put them in, with coefficient $0$.)
But in fact for any $x_0$, the $a_j$ can be computed: they are of the shape $\frac{1}{j!}b_j$, where $b_j$ is one of $\pm \sin(x_0)$ or $\pm \cos(x_0)$. Specifically, $b_0=\sin(x_0)$, $b_1=\cos(x_0)$, $b_2=-\sin(x_0)$, $b_3=-\cos(x_0)$, $b_4=\sin(x_0)$, $b_5=\cos(x_0)$, and continuing to cycle with period $4$ forever.
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